Essential University Physics: Volume 1 (3rd Edition)

by
Wolfson, Richard

Published by
Pearson

ISBN 10:
0321993721

ISBN 13:
978-0-32199-372-4

Chapter 8 - Exercises and Problems - Page 141: 26

Answer

$h=1.7\times 10^6m$

Work Step by Step

We know that according to law of conservation of energy
$\frac{1}{2}mv^2=Gm_Em(\frac{1}{r_1}-\frac{1}{r_2})$
This simplifies to:
$\frac{1}{2}v^2=Gm_E(\frac{1}{r_1}-\frac{1}{r_2})$
We plug in the known values to obtain:
$\frac{1}{2}(5100)^2=6.67\times 10^{-11}(5.97\times 10^{24})(\frac{1}{6.37\times 10^6}-\frac{1}{6.37\times 10^6+h})$
This simplifies to:
$h=1.7\times 10^6m$